CBSE Class 10 Mathematics Standard 2026: question 26

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Step-by-step solution

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1. Suppose, to the contrary, that $\sqrt{3}$ is rational. Then $\sqrt{3}=\frac{p}{q}$, where $p,q$ are integers, $q\ne 0$, and $p,q$ are co-prime (no common factor other than 1).

   ½ mark

2. Squaring, $3=\frac{p^2}{q^2}$, so $p^2=3q^2$.

   Thus 3 divides $p^2$. Since 3 is prime, 3 divides $p$. ...(i)

   1 mark

3. Let $p=3a$ for some integer $a$. Then $9a^2=3q^2\Rightarrow q^2=3a^2$.

   So 3 divides $q^2$, and since 3 is prime, 3 divides $q$. ...(ii)

   1 mark

4. From (i) and (ii), 3 is a common factor of $p$ and $q$. This contradicts the fact that $p$ and $q$ are co-prime.

   The contradiction comes from our assumption, so $\sqrt{3}$ is an irrational number.

   ½ mark

Answer: $\sqrt{3}$ is irrational (proved by contradiction).

Revise the chapter

• Chapter: Real Numbers (notes, formulas, where marks are lost and every past question)
• Topic: Proving Numbers Irrational